我使用邻接矩阵实现图形,但我无法解决分段错误。任何人都可以帮助我指导二维矩阵的动态分配吗?我还想知道2-D数组如何存储在内存中以及如何访问它。
#include<stdio.h>
#include<stdlib.h>
struct Graph{
int V; // To represent number the vertex...
int E; //To represent number the Edge.....
int **Adj; // Two dimensional matrix to form the adjacency matrix...
};
struct Graph *adjMatrixOfGraph(){
int i; //for scanning the edges between them ....
int u,v; // for loop while initliasing the adjacency matrix...
struct Graph *G=(struct Graph*) malloc(sizeof(struct Graph)); //
if(!G){
printf("Memory Error");
return;
}
printf("Number of Vertices");
scanf("%d",&G->V);
printf("%d",G->V);
printf("Number of Edges");
scanf("%d",&G->E);
G->Adj=(int **)malloc(sizeof(G->V * G->V)); //allocating memory for G->Adj);
/*Dynamic memory allocation for Two Dimensional Arrays */
/* G->Adj = malloc(G->V * sizeof(int ));
if(G->Adj == NULL) {
printf( "out of memory\n");
}
for(i = 0; i < G->V; i++) {
G->Adj[i] = malloc(G->V * sizeof(int ));
if(G->Adj[i] == NULL) {
printf( "out of memory\n");
}
}
*/
if(!G->Adj)
{
printf("Memory Error");
return;
}
for(u=0; u < G->V; u++){
for(v=0; v < G->V; v++){
//printf("%d %d",u,v);
G->Adj[u][v]=0; //initalising the complete adjacency matrix to zero.
}
}
//Enter the edges.. and the vertices.
//We are considering this graph as undirected one ...
for(i=0;i< G->E;i++)
{
scanf("Reading Edges %d %d ",&u,&v);
G->Adj[u][v]=1;
G->Adj[u][v]=1;
//if this graph was directed then we should have considere only one side...
//G->V[u][v]=1;
}
return G;
}
main()
{
struct Graph *G1=adjMatrixOfGraph();
//struct Graph *adjMatrixOfGraph(){
printf("Successful");
return 0;
}
答案 0 :(得分:1)
为int **Adj
分配内存的方式如下:
首先,为指向整数的指针数分配内存:
Adj = malloc(sizeof(int*) * number_of_integers); /* notice what I pass to sizeof */
接下来,为每个整数分别分配内存:
for (i = 0; i < number_of_integers; i++)
Adj[i] = malloc(sizeof(int) * G->E);
当然,每个malloc
来电都需要以类似的方式进行free
来电。
请注意,我没有cast the result of malloc。
我对您的代码进行了一些其他更改:
更新scanf
以确保缓冲区中剩余的新行没有问题:
printf("Number of Vertices: ");
scanf(" %d", &G->V);
printf("Number of Edges: ");
scanf(" %d", &G->E);
初始化它们(或者,查找calloc,因为它为你做零初始化):
for(u=0; u < G->V; u++) // for each vertice
{
for(v=0; v < G->E; v++) // for each edge
{
G->Adj[u][v] = 0;
}
}
下面的部分我不确定,你手动将边缘设置为1,对吗?您不应该使用G->V
而不是G->E
吗?
for(i = 0; i < G->V; i++)
{
printf("Reading vertice u: ");
scanf(" %d",&u);
printf("Reading edge v: ");
scanf(" %d",&v);
if (u > G->V || v > G->E) // simple error handling
{
printf("Input bigger than size of vertice/edges\n");
exit(1);
}
G->Adj[u][v] = 1;
G->Adj[u][v] = 1;
}
之后我能够打印Successful
。如果您想更轻松一点,请使用-g
标志编译代码,如果您使用Linux,请执行ulimit -c unlimited
。每次出现段错误时,都会创建一个coredump文件。
然后查看问题所在,运行gdb your_app core
并运行backtrace
。我不能强调在这些情况下使用调试器的重要性。
答案 1 :(得分:1)
int **allocate_2D_array(int rows, int columns)
{
int k = 0;
int **array = malloc(rows * sizeof (int *) );
array[0] = malloc(columns * rows * sizeof (int) );
for (k=1; k < rows; k++)
{
array[k] = array[0] + columns*k;
bzero(array[k], columns * sizeof (int) );
}
bzero(array[0], columns * sizeof (int) );
return array;
}